I'm learning time series analysis and I don't understand why the back-transform of Box-Cox transformation outputs the median instead of the mean of the forecast distribution.
The family of Box-Cox transformations is defined as follows: $$\tag{1} w_t = \begin{cases} \log(y_t) & \text{if $\lambda=0$}; \\ (y_t^\lambda-1)/\lambda & \text{otherwise}. \end{cases}$$
Hence the normal back-transform would be: $$\begin{equation} \tag{2} y_{t} = \begin{cases} \exp(w_{t}) & \text{if $\lambda=0$};\\ (\lambda w_t+1)^{1/\lambda} & \text{otherwise}. \end{cases} \end{equation}$$
In the book that I'm reading "Forecasting: Principles and Practice". It says that:
One issue with using mathematical transformations such as Box-Cox transformations is that the back-transformed point forecast will not be the mean of the forecast distribution. In fact, it will usually be the median of the forecast distribution (assuming that the distribution on the transformed space is symmetric). For many purposes, this is acceptable, but occasionally the mean forecast is required. For example, you may wish to add up sales forecasts from various regions to form a forecast for the whole country. But medians do not add up, whereas means do.
For a Box-Cox transformation, the back-transformed mean is given by:
$$\begin{equation} \tag{3} y_t = \begin{cases} \exp(w_t)\left[1 + \frac{\sigma_h^2}{2}\right] & \text{if $\lambda=0$;}\\ (\lambda w_t+1)^{1/\lambda}\left[1 + \frac{\sigma_h^2(1-\lambda)}{2(\lambda w_t+1)^{2}}\right] & \text{otherwise;} \end{cases} \end{equation}$$
where $\sigma_h^2$ is the $h$-step forecast variance. The larger the forecast variance, the bigger the difference between the mean and the median.
Could you please help me explain why the point forecast in equation (2) is the median and the bias-adjusted forecast in equation (3) is the mean of the forecast distribution and what is the formula for $\sigma_h^2$ in equation (3)?
Many thanks in advance for your help!
The way I understand it is, that the transformation is done so that errors are approximately normal distributed with a stable variance. Hence the error distribution of the transformed data should be symmetric around the forecasted value. When you then back-transform, the back-transformed forecasted error distribution is no longer symmetric. That is, the 'size' of the forecasted back-transformed errors that lie above the back-transformed forecasted (median) value is no longer of the same 'size' as the forecasted back-transformed errors below the back-transformed predicted (median) value. However, there is still the expectation that the number of back-transformed errors that fall above or below the back-transformed predicted value are equal. Hence the back-transformed predicted value is the median of the expected values. For derivation of the mean of the back transformation see: https://robjhyndman.com/hyndsight/backtransforming/